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Position, path length and displacement, average velocity and speed (§2.1 Introduction)
Motion is a change of position with time. But a position number alone means nothing. Two things must be fixed first: an ORIGIN, a reference point, and a POSITIVE DIRECTION along the line.
Take the straight practice track behind a school, marked off in metres from the start line. Everyone reads a runner’s position off the same mark, counting the same way. Move the start line, or flip which way is positive, and every position number on that same, unmoved track changes with it. A position depends entirely on the origin and direction chosen for it, and means nothing until both are fixed. A runner three metres before the start line is at $x = -3 m$ only once the near side is agreed to be negative.
The common slip is treating a position, or its sign, as a fact about the runner. It is really a fact about the convention chosen in advance. A negative position is not strange or special. It is only what a point on the near side of the origin is called, once a direction has been agreed.
One more idealisation is worth noting here. Straight-line motion treats the runner as a POINT OBJECT, its own size ignored against the distance it covers. That is honest for a sprinter on a hundred-metre track. It stops being honest only once the measured distance shrinks down toward the runner’s own size.
PATH LENGTH is the total length of track actually covered. Every metre counts, whichever direction it was run in. DISPLACEMENT asks a different question: how far, and which way, did the runner end up from the start? It is the signed change $Delta x = x_2 - x_1$, using only the two endpoints — not one metre of the road in between.
Send a runner sixty metres down the same practice track, then straight back the whole sixty, to cool down. Path length is a plain sum: $60 + 60 = 120$ metres, every stride counted. Displacement holds only the two endpoints apart. The finish, $x_2$, sits exactly where the start, $x_1$, did. So $Delta x = 0$. Displacement can shrink to zero exactly when path length is at its largest for that trip — the two measure completely different things.
Treating ‘distance’ and ‘displacement’ as two names for one number is the trap this idea corrects. They coincide only for a run that never reverses: an out-and-back sprint straight to the sixty-metre mark, and no further, gives distance and $|Delta x|$ both equal to sixty. The moment the runner turns around even once, $|Delta x|$ falls below the path length — and stays there. In general, $|Delta x| \leq$ path length, always.
AVERAGE VELOCITY over an interval is the signed ratio $v_\text{avg} = Delta x/Delta t = (x_2 - x_1)/(t_2 - t_1)$. It reports which way the runner’s position shifted overall, and how fast, across the whole interval. AVERAGE SPEED is different: path length divided by $Delta t$, always positive, since path length itself is never negative.
A coach timing the sixty-out-sixty-back cool-down gets an average velocity of exactly zero. The runner ended up back at the start, whatever pace was run in between. The stopwatch’s other number refuses to go anywhere near zero: 120 metres covered, over however many seconds the whole run took. *Because path length is never smaller than $|Delta x|$, average speed is never smaller than $|v_\text{avg}|$* — equal only for a trip that never turns back.
Average velocity is a verdict on the WHOLE interval. It says nothing about any single instant inside it. Two runners covering the same sixty metres in twelve seconds share one average velocity. It makes no difference whether one held a steady pace, or sprinted first and walked later.
Instantaneous velocity and speed (§2.2)
A runner’s motion can be read straight off a graph. Plot the metre mark on one axis and the stopwatch reading on the other. The SLOPE of the CHORD joining any two points on that curve equals the average velocity over that interval — no extra arithmetic needed.
One common misreading treats the HEIGHT of the graph as the velocity. Height is only ever position — where the runner was, not how fast they were moving there. A second misreading treats a steeper stretch as meaning the runner went further. Steeper only ever means faster. Two runners on the same graph, covering the same sixty metres on lines of different steepness, have run the identical distance in different times.
The chord’s slope is only ever an average over the stretch it spans. It says nothing about what happened partway through that stretch. Shrink the interval, sliding the chord’s two points toward each other along the curve, and the chord pivots. It settles ever closer against the curve, at a single instant. That pivoting chord sets up the next idea: instantaneous velocity.
Keep shrinking the interval around one instant on the x-t graph. The chord’s slope does not wander. It settles on one particular number and stays there. That settled number is INSTANTANEOUS VELOCITY, written $v = (d x)/(d t)$.
Build this numerically first, before trusting the symbol. Take smaller and smaller windows around one instant: one second wide, then a tenth, then a hundredth. Compute $Delta x/Delta t$ fresh each time. The ratio does not blow up or vanish. It settles on one steady value. Do the same with a stopwatch at the forty-metre mark: splits taken closer and closer together converge on one number too. That number is the runner’s speed AT that mark — not their speed over the whole race.
Here is the trap the symbol $(d x)/(d t)$ invites. It is NOT an ordinary fraction that splits apart into a separate $d x$ and a separate $d t$, however it looks on the page. The symbol names the LIMIT a ratio settles on — it is not a fraction any calculation is free to break in half. Nor is it the same computation as average velocity’s total-distance-over-total-time.
Graphically, $(d x)/(d t)$ is the slope of the TANGENT to the x-t curve at one instant. It is the limit of the very chord from k4, pivoting until its two points merge into one.
INSTANTANEOUS SPEED is the size of instantaneous velocity, direction stripped away. A runner clocked at $+7 m/s$ and one clocked at $-7 m/s$, running opposite ways on the same track, share exactly the same speed: 7 m/s.
The average case allowed a gap. Average speed could run ahead of $|v_\text{avg}|$ whenever the runner reversed somewhere inside the interval. The sixty-out-sixty-back cool-down proves it: an average speed of several metres per second, against an average velocity of exactly zero. *No such gap survives at a single instant: instantaneous speed equals $|v|$ exactly, every time.* One instant leaves no room inside it for a reversal to hide.
Acceleration — average, instantaneous, and the x-t/v-t graph pair (§2.3)
ACCELERATION measures how fast VELOCITY itself changes. Average acceleration over an interval is $a_\text{avg} = Delta v/Delta t$. Instantaneous acceleration is $a = (d v)/(d t)$ — the slope of the velocity-time graph, built the same way instantaneous velocity (k5) was.
Throw a cricket ball straight up over a school compound wall. Watch it the whole way: rising, slowing, momentarily still at the very top, then falling back, speeding up all over again. Through every one of those different-looking stretches, the ball’s acceleration never changes. It is $a = -g$ the entire time — going up, at the top, and coming down.
Here is the trap that scene sets. The ball looks still at the top, so it is tempting to say nothing is changing there, and that acceleration must be zero too. *Neither is true. Velocity is exactly zero at the top, yet the full acceleration $g$ is still acting downward.* That is exactly why the ball does not hang in the air. It falls straight back.
The SIGN of acceleration only ever gives its DIRECTION. Whether a body speeds up or slows down depends on whether that acceleration acts WITH the velocity or AGAINST it. This was Galileo’s own point, from his study of free fall: acceleration tracks velocity’s change with TIME, not with distance covered.
Put the two graphs of one motion side by side. For UNIFORM velocity, the x-t graph is a straight slanted line and the v-t graph is flat. For UNIFORM ACCELERATION, it flips: the v-t graph becomes the straight slanted line, its slope equal to $a$, and the x-t graph curves into a parabola.
The v-t graph hides one more fact. The AREA under the curve, between two times, equals the DISPLACEMENT over that interval. It adds up every instant’s own contribution — k5’s derivative, run in reverse. Read a coach’s velocity-time plot of the cool-down jog, and the shaded region between two stopwatch marks gives the metres covered directly, no separate calculation needed.
The commonest misreading treats that shaded area as the SPEED or the ACCELERATION, rather than the displacement it actually is. A related one reads a parabola’s changing curvature as a change in the TYPE of motion. It is only the steepening RATE of the same, uniformly accelerating motion. Area below the time axis subtracts from the running total, rather than adding to it. Picture a runner who overshoots the finish line and steps back. That backward stretch covers area below the axis. It comes off the total distance, not onto it.
Kinematic equations for uniformly accelerated motion, free fall, Galileo's odd numbers (§2.4)
For CONSTANT acceleration, three equations between them solve almost every straight-line problem. The first, $v = v_0 + a t$, comes directly from holding $a = Delta v/Delta t$ fixed. The second, $x = v_0 t + (1/2) a t^2$, is the area under the v-t line from k8. It splits into a rectangle of height $v_0$ and a triangle of height $a t$. The third, $v^2 = v_0^2 + 2 a x$, is what is left once $t$ is eliminated between the first two.
A train pulls out of a station with a steady acceleration and a starting speed $v_0$. That is enough to check any one of five quantities — $v_0, v, a, t, x$ — as soon as the other four are known. Pick whichever equation already contains them and leaves out the one thing neither known nor wanted.
Starting not at the platform’s origin, but some distance $x_0$ down the track, changes nothing about the method. Replace $x$ by $(x - x_0)$ throughout, and the same three equations run exactly as before. None of the three equations survive once acceleration itself stops being constant. A train that brakes partway through the interval breaks that condition — and none of the three equations still apply.
FREE FALL is motion under gravity alone. It hides a genuinely surprising fact: near the Earth’s surface, every object accelerates downward at the identical rate, $g = 9.8 m/s^2$, whatever its mass. Chapter 4’s Second Law explains later why mass cancels out. Here, it is simply what a falling coin and a falling brick both do, measured.
Drop a coin and a crumpled ball of paper of about the same size, from a rooftop, at the same instant. Set aside the everyday guess for a moment. They land together. A heavier object does not fall faster than a lighter one — mass plays no part at all in free-fall acceleration. Only a sheet spread flat, instead of crumpled, visibly lags. That lag comes from air resistance pushing against its shape, not from gravity treating the two objects differently.
Taking upward as positive, free fall becomes uniformly accelerated motion with $a = -g$. So k9’s three kinematic equations apply completely unchanged. They carry $a = -g$ throughout, including at the highest point of a ball thrown up. There, velocity is momentarily zero (k7), but the acceleration is still the full $g$ downward. A consistent sign convention for up and down, held before a single number goes in, is what keeps a free-fall calculation honest from the first line to the last.
Free fall from rest hides a pattern inside its own equation. The distance covered by time $t$ is $x = (1/2) g t^2$. Look at the distance fallen in the first, second, third — and every later — EQUAL time interval. Those distances run in a fixed ratio: $1, 3, 5, 7$, and onward through the odd numbers.
Mark equal time intervals with a metronome. Photograph a stone falling past a ruled scale at each tick. The gaps between successive marks do not stay equal — they grow, and they grow in exactly that $1:3:5:7$ pattern. This is the direct signature of distance going as $t^2$. It was Galileo’s own experimental fingerprint of uniform acceleration, read off centuries before ‘acceleration’ had a formal definition at all.
The standard slip is expecting equal distances in equal times. That is a fair guess — but it only describes UNIFORM VELOCITY. Under uniform ACCELERATION, the gaps between marks have to grow. The falling body is moving faster by the end of each successive second than it was at the end of the one before.
Equal times, growing gaps — that is the whole signature of acceleration. A student who can read it off a ruled photograph the night before an exam has understood free fall properly.
Relative velocity (§2.5)
This last idea sits outside the CBSE syllabus. The current NCERT reprint has dropped it too. Nothing here will be examined. It earns a place anyway, because the physics is real. Anyone who has watched one train pass or catch up with another already carries most of the intuition before a single equation appears.
Velocity is never measured in a vacuum. It is always measured RELATIVE TO something, usually the ground, though the frame is a free choice. The same motion looks different depending on which frame is chosen. Object A moves with velocity $v_A$. Object B moves with velocity $v_B$. Both are read off the same frame, along the same line. The velocity of A RELATIVE TO B is $v_\text{AB} = v_A - v_B$ — exactly the velocity A appears to have, seen from inside B.
Two trains glide along parallel tracks at the identical speed, in the same direction. Each passenger sees the other train as apparently parked: $v_\text{AB} = 0$. Yet both trains move at full speed relative to the platform. Let the two trains approach on facing tracks instead, and the relative velocities do not cancel. They combine, since subtracting a negative speed adds it — exactly why a head-on closing between two trains is so much more violent than one catching up on another from behind.